Question

B= Balok
Ba => Bb => Bc => F=145N
Jika koefisien gesek antar balok A, B, dan C tehadap bidang datar masing masing 0,1 0,2 dan 0,3.
Massa A=10 kg, B=15 kg, C=10 kg .Besar Gaya F=145N
.tentukan:
a) Percepatan
B) Tegangan Tali BC
c) Tegangan tali AB

Answer 1

Verified answer

FgesA = μA.mA.g = 0,1 . 10 . 10 = 10 Newton

FgesB = μB.mB.g = 0,2 . 15 . 10 = 30 Newton

FgesC = μC.mC.g = 0,3 . 10 . 10 = 30 Newton

a)
ΣF = mtotal . a
F - FgesA - FgesB - FgesC = (10+15+10) . a
145 - 10 - 30 - 30 = 35.a
75 = 35a
a = 2 m/s^2

b) tegangan tali antara benda B dan C
Tegagan tali B C menarik benda A dan B
ΣF = (mA+mB).a
Tbc - FgesA - FgesB = (10+15).2
Tbc - 10 - 30 = 25 . 2
Tbc - 40 = 50
Tbc = 90 N

c) tegangan tali diantara benda A dan B menarik benda A saja
ΣF = mA. a
Tab - FgesA = 10.2
Tab - 10 = 20
Tab = 30 Newton